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Chapter 9
Gases: Their Properties and Behavior
Properties of Gases
• There are 5 important properties of gases:
– Confined gases exerts pressure on the wall of a
container uniformly
– Gases have low densities
– Gases can be compressed
– Gases can expand to fill their contained uniformly
– Gases mix completely with other gases in the same
container
Chapter 9
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Kinetic Molecular Theory of Gases
•
The Kinetic Molecular Theory of Gases is the model used
to explain the behavior of gases in nature.
•
This theory presents physical properties of gases in terms of
the motion of individual molecules:
1. Average Kinetic Energy  Kelvin Temperature
2. Gas molecules are points separated by a great distance
3. Particle volume is negligible compared to gas volume
4. Gas molecules are in rapid random motion
5. Gas collisions are perfectly elastic
6. Gas molecules experience no attraction or repulsion
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Properties that Describe a Gas
• These properties are all related to one another.
• When one variable changes, it causes the other
three to react in a predictable manner.
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Gas Pressure (P)
• Gas pressure (P) is the result of constantly moving
gas molecules striking the inside surface of their
container.
Chapter 9
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Atmospheric Pressure
• Atmospheric pressure is the pressure exerted by the air
on the earth.
• Evangelista Torricelli invented
the barometer in 1643 to
measure atmospheric pressure.
• Atmospheric pressure is 760
mm of mercury or 1
atmosphere (atm) at sea level.
What happens to the atmospheric
pressure as you go up in elevation?
Chapter 9
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Measurement of Gas Pressure
• Traditionally, the gas pressure inside of a container is
measured with a manometer
Chapter 9
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Units of Pressure
• Standard pressure is the atmospheric pressure at
sea level, 760 mm of mercury.
– Here is standard pressure expressed in other units:
Chapter 9
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Gas Pressure Conversions
• The barometric pressure is 697.2 torr. What is the
barometric pressure in atmospheres? In mm Hg? In
Pascals (Pa)?
Chapter 9
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Gas Law Problems
Chapter 9
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Boyle’s Law (V and P)
• Boyle’s Law states that the volume of a gas is inversely
proportional to the pressure at constant temperature.
• Mathematically, we write:
P 1.
V
• For a before and after
situation:
P1V1 = P2V2
Chapter 9
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Boyle’s Law Problem
• A 1.50 L sample of methane gas exerts a pressure of
1650 mm Hg. What is the final pressure if the
volume changes to 7.00 L?
P
V
1
1
= P2
P1V1 = P2V2 rearranges to
V2
(1650 mm Hg )(1.50 L)
= 354 mm Hg
7.00 L
Chapter 9
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Charles’ Law (V and T)
• In 1783, Jacques Charles discovered (while hot air ballooning) that
the volume of a gas is directly proportional to the temperature in
Kelvin.
• Mathematically, we write:
TV
• For a before and after
situation:
V1
=
T1
Chapter 9
V2
T2
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Charles’ Law Problem
• A 275 L helium balloon is heated from 20C to 40C.
What is the final volume at constant P?
V1
V2
=
T1
T2
V1 T2
= V2
rearranges to
T1
(275 L)(313 K)
= 294 L
293 K
Chapter 9
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Gay-Lussac’s Law (P and T)
• In 1802, Joseph Gay-Lussac discovered that the pressure of a
gas is directly proportional to the temperature in Kelvin.
• Mathematically, we write:
TP
• For a before and after situation:
P1
=
T1
Chapter 9
P2
T2
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Gay-Lussac’s Law Problem
• A steel container of nitrous oxide at 15.0 atm is cooled
from 25C to –40C. What is the final volume at
constant V?
P1 T2
P1
P2
= P2
rearranges to
=
T1
T1
T2
(15.0 atm)(298 K)
= 11.7 atm
233 K
Chapter 9
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Avogadro’s Law (n and V)
• In the previous laws, the amount of gas was always constant.
• However, the amount of a gas (n) is directly proportional to the
volume of the gas, meaning that as the amount of gas increases,
so does the volume.
• Mathematically, we write:
nV
• For a before and after situation:
V1
=
n1
Chapter 9
V2
n2
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Avogadro’s Law Problem
• A steel container contains 2.6 mol of nitrous oxide with a volume
15.0 L. If the amount of nitrous oxide is increased to 8.4 mol, what
is the final volume at constant T and P?
V1
V2
V1 n 2
rearranges to
= V2
=
n1
n1
n2
(15.0 L)(8.4 mol)
= 48.5 L
2.6 mol
Chapter 9
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Combined Gas Law
• When we introduced Boyle’s, Charles’, and Gay-Lussac’s
Laws, we assumed that one of the variables remained
constant.
• Experimentally, all three (temperature, pressure, and
volume) usually change.
• By combining all three laws, we obtain the combined gas
law:
P1V1 P2V2
=
T1
T2
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Combined Gas Law Problem
• Oxygen gas is normally sold in 49.0 L steel containers
at a pressure of 150.0 atm. What volume would the
gas occupy if the pressure was reduced to 1.02 atm
and the temperature raised from 20oC to 35oC?
Chapter 9
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Molar Volume and STP
• Standard temperature and pressure (STP) are defined as 0C and 1 atm.
• At standard temperature and pressure, one mole of any gas occupies 22.4 L.
• The volume occupied by one mole of gas (22.4 L) is called the molar volume.
1 mole Gas = 22.4 L
Chapter 9
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Molar Volume Calculation – Volume to Moles
• A sample of methane, CH4, occupies 4.50 L at STP. How
many moles of methane are present?
Volume
4.50 L CH4 ×
Chapter 9
Molar
Volume
Moles
1 mol CH4
= 0.201 mol CH4
22.4 L CH4
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Mole Unit Factors
• We now have three interpretations for the mole:
1 mol = 6.02 × 1023 particles
1 mol = molar mass
1 mol = 22.4 L (at STP for a gas)
• This gives us 3 unit factors to use to convert
between moles, particles, mass, and volume.
Chapter 9
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Mole Calculation - Grams to Volume
• What is the mass of 3.36 L of ozone gas, O3, at STP?
Grams
Molar
Mass
Moles
Molar
Volume
Volume
1 mol O3
48.00 g O3
3.36 L O3 ×
×
= 7.20 g O3
22.4 L O3
1 mol O3
Chapter 9
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Mole Calculation – Molecules to Volume
• How many molecules of hydrogen gas, H2,
occupy 0.500 L at STP?
Volume
Molar
Volume
Moles
Avogadro’s
Number
Atoms
1 mol H2
6.02×1023 molecules H2
×
0.500 L H2 ×
22.4 L H2
1 mole H2
1.34 × 1022 molecules H2
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Gas Density and Molar Mass
• The density of a gas is much less than that of a
liquid.
• We can calculate the density of any gas at STP
easily.
molar mass in grams (MM)
= density, g/L
molar volume in liters (MV)
• You can rearrange this equation to find the
Molar mass of an unknown gas too!
Chapter 9
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Calculating Gas Density
• What is the density of ammonia gas, NH3, at STP?
17.04 g/mol
22.4 L/mol
= 0.761 g/L
• 1.96 g of an unknown gas occupies 1.00 L at STP.
What is the molar mass?
Chapter 9
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The Ideal Gas Law
•
When working in the lab, you will not always be at STP.
•
The four properties used in the measurement of a gas (Pressure,
Volume, Temperature and moles) can be combined into a single
gas law:
PV = nRT
•
Here, R is the ideal gas constant and has a value of:
0.0821 atmL/molK
•
Note the units of R. When working problems with the Ideal Gas
Law, your units of P, V, T and n must match those in the constant!
Chapter 9
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Ideal Gas Law Problem
• Sulfur hexafluoride (SF6) is a colorless,
odorless, very unreactive gas. Calculate the
pressure (in atm) exerted by 1.82 moles of
the gas in a steel vessel of volume 5.43 L at
69.5°C.
Chapter 9
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Ideal Gas Law and Molar Mass
• Density and Molar Mass Calculations:
mass
nM PM
d


volume
V
R T
• You can calculate the density or molar mass (M) of a
gas.
• The density of a gas is usually very low under
atmospheric conditions.
Chapter 9
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Ideal Gas Law and Molar Mass
• What is the molar mass of a gas with a density of 1.342
g/L–1 at STP?
• What is the density of uranium hexafluoride, UF6, (MM =
352 g/mol) under conditions of STP?
• The density of a gaseous compound is 3.38 g/L–1 at 40°C
and 1.97 atm. What is its molar mass?
Chapter 9
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Gases in Chemical Reactions
•
Gases are involved as reactants
and/or products in numerous
chemical reactions.
•
Typically, the information given for
a gas in a reaction is its Pressure
(P), volume (V) (or amount of the
gas (n)) and temperature (T).
•
We use this information and the
Ideal Gas Law to determine the
moles of the gas (n) or the volume
of the gas (V).
•
Once we have this information, we
can proceed with the problem as we
would any other stoichiometry
problem.
Chapter 9
A (g) + X (s) → B (s) + Y (l)
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Reaction with a Gas
• Hydrogen gas is formed when zinc metal reacts
with hydrochloric acid. How many liters of
hydrogen gas at STP are produced when 15.8 g of
zinc reacts?
Zn (s) + 2HCl (aq) → H2 (g) + ZnCl2 (aq)
Grams of Zn
Molar
Mass of Zn
Moles of Zn
Mole
Ratio
Moles of Hh
Molar
Volume
Liters of H2
Can use because at STP!
Chapter 9
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Reaction with a Gas
• Hydrogen gas is formed when zinc metal reacts with
hydrochloric acid. How many liters of hydrogen gas at a
pressure of 755 atm and 35°C are produced when 15.8 g
of zinc reacts?
Zn (s) + 2HCl (aq) → H2 (g) + ZnCl2 (aq)
Grams of Zn
Molar
Mass of Zn
Moles of Zn
Mole
Ratio
Moles of Hh
Ideal Gas
Law
Liters of H2
Use because not at STP!
Chapter 9
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Dalton’s Law of Partial Pressures
• In a mixture of gases the total pressure, Ptot, is the sum
of the partial pressures of the gases:
Ptot = P1 + P2 + P3 + etc.
RT
P total 
V
n
• Dalton’s law allows us to work with mixtures of gases.
Chapter 9
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Dalton’s Law of Partial Pressures
• For a two-component system, the moles of
components A and B can be represented by the
mole fractions (XA and XB).
XA 
nA
nA + nB
XB 
nB
nA + nB
XA + XB  1
• What is the mole fraction of each component in
a mixture of 12.45 g of H2, 60.67 g of N2, and
2.38 g of NH3?
Chapter 9
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Dalton’s Law of Partial Pressures
• Mole fraction is related to the total pressure by:
Pi  X i Ptot
• On a humid day in summer, the mole fraction of
gaseous H2O (water vapor) in the air at 25°C can be
as high as 0.0287. Assuming a total pressure of
0.977 atm, what is the partial pressure (in atm) of
H2O in the air?
Chapter 9
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Dalton’s Law of Partial Pressures
• Exactly 2.0 moles of Ne and 3.0 moles of Ar were
placed in a 40.0 L container at 25°C. What are the
partial pressures of each gas and the total pressure?
• A sample of natural gas contains 6.25 moles of
methane (CH4), 0.500 moles of ethane (C2H6), and
0.100 moles of propane (C3H8). If the total pressure
of the gas is 1.50 atm, what are the partial pressures
of the gases?
Chapter 9
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Kinetic Molecular Theory of Gases
•
The Kinetic Molecular Theory of Gases is the
model used to explain the behavior of gases in
nature.
•
This theory presents physical properties of gases in
terms of the motion of individual molecules.
1. Average Kinetic Energy  Kelvin Temperature
2. Gas molecules are points separated by a great distance
3. Particle volume is negligible compared to gas volume
4. Gas molecules are in rapid random motion
5. Gas collisions are perfectly elastic
6. Gas molecules experience no attraction or repulsion
Chapter 9
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Kinetic Molecular Theory of Gases
Chapter 9
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Kinetic Molecular Theory of Gases
• Average Kinetic Energy (KE) is given by:
1 2 3RT
KE  mu 
2
2 NA
3RT 3RT
2
u 

mNA MM
Chapter 9
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Kinetic Molecular Theory of Gases
• The Root–Mean–Square Speed (uRMS): is a measure
of the average molecular speed of a particle of gas.
3RT
u 
MM
2
Taking square root of both
sides gives the equation
u rms
Chapter 9
3RT

MM
R = 8.314 J/mol K
42
Kinetic Molecular Theory of Gases
• Calculate the root–mean–square speeds (uRMS)
of helium atoms and nitrogen molecules in m/s
at 25°C.
Chapter 9
43
Graham’s Law: Diffusion and Effusion
• Diffusion is the mixing
of different gases by
random molecular
motion and collision.
Chapter 9
• Effusion is when gas
molecules escape without
collision, through a tiny
hole into a vacuum.
44
Graham’s Law: Diffusion and Effusion
• Graham’s Law: The rate of effusion is
proportional to its RMS speed (uRMS).
3RT
Rate  u rms 
MM
• For two gases at same temperature and
pressure:
MM 2
Rate1
MM 2


Rate2
MM 1
MM 1
Chapter 9
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Graham’s Law: Diffusion and Effusion
• Under the same conditions, an unknown gas diffuses
0.644 times as fast as sulfur hexafluoride, SF6 (MM =
146 g/mol). What is the identity of the unknown gas
if it is also a hexafluoride?
• What are the relative rates of diffusion of the three
naturally occurring isotopes of neon: 20Ne, 21Ne, and
22Ne?
Chapter 9
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Behavior of Real Gases
•
Deviations from Ideal behavior result from
two key assumptions about ideal gases.
1. Molecules in gaseous state do not exert any force,
either attractive or repulsive, on one another.
2. Volume of the molecules is negligibly small compared
with that of the container.
•
These assumptions breakdown at high
pressures, low volumes and low
temperatures.
Chapter 9
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Behavior of Real Gases
• At STP, the volume occupied by a single molecule is very small relative to its
share of the total volume
–
For example, a He atom (radius = 31 pm) has roughly the same space to move
about as a pea in a basketball
• Let’s say we increase the pressure of the system to 1000 atm, this will cause
a decrease in the volume the gas has to move about in
–
Now our He atom is like a pea in a ping pong ball
• Therefore, at high pressures, the volume occupied by the gaseous molecules
is NOT negligible and must be considered.
• So the space the gas has to move around in is less than under Ideal
conditions!
VReal > VIdeal
Chapter 9
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Behavior of Real Gases
• At low volumes, particles are much closer together and
attractive forces become more important than at high volumes.
• This increase in intermolecular attractions pulls the molecules
away from the walls of the containers, meaning that they do not
hit the wall with as great a force, so the pressure is lower than
under ideal conditions.
Chapter 9
PReal < PIdeal
49
Behavior of Real Gases
• A similar phenomenon is seen at low temperatures
(aka. The Flirting Effect)
–
As molecules slow down, they have more time to interact
therefore increasing the effect of intermolecular forces.
• Again, this increase in intermolecular attractions pulls
the molecules away from the walls of the containers,
meaning that they do not hit the wall with as great a
force, so the pressure is lower than under ideal
conditions.
Chapter 9
PReal < PIdeal
50
Behavior of Real Gases
Chapter 9
51
Behavior of Real Gases
• Corrections for non-ideality require the van der
Waals equation.
an

P + 2
V

2
Correction for
Intermolecular
Attractions
PReal < PIdeal
Chapter 9

  V – nb   nRT

Correction for
VReal > VIdeal
Molecular Volume
n = moles of gas
a and b are constants given in the problem
52
Behavior of Real Gases
• Given that 3.50 moles of NH3 occupy 5.20 L at 47°C,
calculate the pressure of the gas (in atm) using:
(a) the ideal gas equation
(b) the van der Waals equation. (a = 4.17, b = 0.0371)
• Calculate the pressure exerted by 4.37 moles of molecular
chlorine confined in a volume of 2.45 L at 38°C.
Compare the pressure with that calculated using the ideal
gas equation. (a = 6.49 and b = 0.0562)
Chapter 9
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